5) Two point charges of +20.0 μC and -8.00 μC are separated by a distance of...

5) Two point charges of +20.0 μC and -8.00 μC are separated by a distance of 20.0 cm. What is the intensity of electric field E midway between these two charges?

a. 25.2 × 10⁵ N/C directed towards the positive charge

b. 25.2 × 10⁶N/C directed towards the positive charge

c. 25.2 × 10⁴ N/C directed towards the negative charge

d. 25.2 × 10⁵ N/C directed towards the negative charge

e. 25.2 × 10⁶ N/C directed towards the negative charge

Expert Solution

The electric intensity at the mid point is equal to thealgebraic sum of the intensities due to individual charges.
         i.e., E= E1 + E2 = k(q1 + q2)/r2         where k = (1/4πε) = 8 * 10^9 for freespace.
                                                              andsince the distance from the two points is same.
             Hence E = 8 * 10^9 ( 20 + 8 ) * 10^-6 / (10 *10^-2)2
      On simplifying weget,   E = 3.51 * 10^7N/c
Hope this helps you.

genius_generous answered 2 years ago

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